Is cot(x)(1-cos^2(x))= 1/2 sin(2x) ?

The answer to whether cot(x)(1-cos^2(x))= 1/2 sin(2x) is True

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prove cot(x)(1-cos^2(x))= 1/2 sin(2x)

prove

cot (x) (1 - cos^{2} (x)) = \frac{1}{2} sin (2 x)

True

Solution steps

cot (x) (1 - cos^{2} (x)) = \frac{1}{2} sin (2 x)

Manipulating left side

cot (x) (1 - cos^{2} (x))

Express with sin, cos

(1 - cos^{2} (x)) cot (x)

Use the basic trigonometric identity:

cot (x) = \frac{cos ( x )}{sin ( x )}

= (1 - cos^{2} (x)) \frac{cos ( x )}{sin ( x )}

Simplify

(1 - cos^{2} (x)) \frac{cos ( x )}{sin ( x )} : \frac{cos ( x ) ( 1 - cos ^{2} ( x ) )}{sin ( x )}

(1 - cos^{2} (x)) \frac{cos ( x )}{sin ( x )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{cos ( x ) ( 1 - cos ^{2} ( x ) )}{sin ( x )}

= \frac{cos ( x ) ( 1 - cos ^{2} ( x ) )}{sin ( x )}

= \frac{( 1 - cos ^{2} ( x ) ) cos ( x )}{sin ( x )}

Rewrite using trig identities

\frac{( 1 - cos ^{2} ( x ) ) cos ( x )}{sin ( x )}

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

1 - cos^{2} (x) = sin^{2} (x)

= \frac{sin ^{2} ( x ) cos ( x )}{sin ( x )}

Cancel the common factor:

sin (x)

= sin (x) cos (x)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

sin (x) cos (x) = \frac{sin ( 2 x )}{2}

= \frac{sin ( 2 x )}{2}

= \frac{sin ( 2 x )}{2}

Manipulating right side

\frac{1}{2} sin (2 x)

Simplify

= \frac{sin ( 2 x )}{2}

We showed that the two sides could take the same form

\Rightarrow True

p ro v e \frac{sec ^{2} ( t )}{tan ( t )} = sec (t) csc (t)

p ro v e cos (x) (tan^{2} (x) + 1) = sec (x)

p ro v e csc^{2} (θ) - cot^{2} (θ) = 1

p ro v e (sec^{2} (x) - 1) csc^{2} (x) = sec^{2} (x)

p ro v e (1 + sin (a)) (1 - sin (a)) = cos^{2} (a)

Is cot(x)(1-cos^2(x))= 1/2 sin(2x) ?
The answer to whether cot(x)(1-cos^2(x))= 1/2 sin(2x) is True